\[\boxed{\mathbf{1118.ОК\ ГДЗ - домашка\ на}\ 5}\mathbf{\ }\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[d_{основ} = 6,6\ м.\]
\[\mathbf{Найти:}\]
\[S_{основ} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S = \pi R^{2}.\]
\[2)\ R = \frac{d}{2} = \frac{6,6}{2} = 3,3.\]
\[3)\ S = 3,14 \bullet {3,3}^{2} =\]
\[= 3,14 \bullet 10,89 \approx 34,19\ м^{2}.\]
\[Ответ:площадь\ основания\ \]
\[колокола\ равна\ 34,19\ м^{2}.\]
\[\boxed{\mathbf{1118.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[BC = a;\ \]
\[\angle B = \alpha;\]
\[\angle C = \beta.\]
\[\mathbf{Найти:}\]
\[CC_{1};BB_{1};AA_{1} - биссектрисы.\]
\[\mathbf{Решение.}\]
\[1)\ Рассмотрим\ \mathrm{\Delta}BB_{1}C:\]
\[\angle BB_{1}C = 180{^\circ} - \frac{\alpha}{2} - \beta.\]
\[По\ теореме\ синусов:\]
\[\frac{\text{BC}}{\sin{\angle BB_{1}C}} = \frac{BB_{1}}{\sin\beta}\]
\[BB_{1} = \frac{{a \bullet \sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)}.\]
\[2)\ Рассмотрим\ \mathrm{\Delta}BCC_{1}:\]
\[\angle BC_{1}C = 180{^\circ} - \alpha - \frac{\beta}{2}.\]
\[По\ теореме\ синусов:\]
\[\frac{\text{BC}}{\sin{\angle BC_{1}C}} = \frac{CC_{1}}{\sin\alpha}\]
\[CC_{1} = \frac{{a \bullet \sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)}.\]
\[3)\ Рассмотрим\ \mathrm{\Delta}BAA_{1}:\]
\[\angle BAA_{1} = 90{^\circ} - \frac{\beta + \alpha}{2};\ \]
\[\angle BA_{1}A = 90{^\circ} + \frac{\beta - \alpha}{2}.\]
\[По\ теореме\ синусов:\]
\[\frac{AA_{1}}{\sin{\angle B}} = \frac{\text{AB}}{\sin{\angle A_{1}}}\]
\[AA_{1} = \frac{AB \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)};\]
\[AB = \frac{a \bullet \sin\beta}{\sin(\alpha + \beta)};\]
\[AA_{1} = \frac{\frac{a \bullet \sin\beta}{\sin(\alpha + \beta)} \bullet \sin\alpha}{\sin\left( 90{^\circ} + \frac{\beta - \alpha}{2} \right)} =\]
\[= \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]
\[Ответ:CC_{1} = \frac{{a \bullet \sin}\alpha}{\sin\left( \alpha + \frac{\beta}{2} \right)};\ \]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B}B_{1} = \frac{{a \bullet \sin}\beta}{\sin\left( \beta + \frac{\alpha}{2} \right)};\]
\[AA_{1} = \frac{a \bullet \sin\alpha \bullet \sin\beta}{\sin(\alpha + \beta) \bullet \cos\left( \frac{\beta - \alpha}{2} \right)}.\]